Let $X=(X_n)_{n\in\mathbb N}$ be a stochastic process in $\{0,1\}^{\mathbb N}$ with distribution $\mu$. I do not at first make any assumptions about $X$ being stationary or having any kind of correlations or associations. Let $Y=(Y_n)_{n\in\mathbb N}$ be an iid Bernoulli process with distribution product measure $\pi_p$, i.e. $\mathbb P(Y_n=1)=p$ for all $n$ and independently of all other $Y_j$.
I want to study the conditions under which $\pi_p\preceq \mu$, that $X$ stochastically dominates $Y$. The order $X(\omega)\geq Y(\omega)$ uses the natural ordering of their sample paths for each $\omega\in\Omega$ a common probability space upon which they are constructed, for which I just say $X\geq Y$.
From Lemma 1.1 in Liggett et al. (1997), I can conclude that a sufficient condition for $\pi_p\preceq \mu$ is that $$ \mathbb P(X_n=1\mid X_{s_j}=\epsilon_{s_j}, \text{ for all } j=1,2,\ldots,m)\geq p \tag{1}\label{eq1} $$ for any $n$ and any $\{s_1,s_2,\ldots,s_m\}\subset \{1,2,\ldots,n-1\}$ and any $\epsilon=(\epsilon_n)_{n\in\mathbb N}$. Of course we must be conditioning on an event of positive probability, so I always assume that.
My question is on why this condition isn't also necessary.
I know that $\pi_p\preceq \mu$ if and only if there exists a coupling of $X$ and $Y$ such that $\mathbb P(X\geq Y)=1$. (I hope it is clear that I am now talking about coupled "copies of" $X$ and $Y$ with the appropriate marginals.) Such a coupling must also satisfy (\ref{eq1}) yes?
I use $\mathbb P^\epsilon$ to refer to the conditional joint distribution of $(X,Y)$, conditioned on $X_{s_j}=\epsilon_{s_j}$ for $j=1,2,\ldots,m$: $$\mathbb P^\epsilon(X_n=1)=\mathbb P^\epsilon(X_n=1,Y_n=0)+\mathbb P^\epsilon(X_n=1,Y_n=1)$$ Since the coupling satisfies $\mathbb P(X\geq Y)=1$, we also get $\mathbb P^\epsilon(X\geq Y)=1$ (assuming we condition on an event of positive probability). Hence $\mathbb P^\epsilon(X_n=1,Y_n=1)=\mathbb P^\epsilon(Y_n=1)=p$ (since $\{X_n=0,Y_n=1\}$ has probability zero under $\mathbb P$ and $\mathbb P^\epsilon$), therefore we clearly have $\mathbb P^\epsilon(X_n=1)\geq p$ as desired.
So I have (\ref{eq1}, for all $n,\epsilon,m$) if and only if $\pi_p\preceq\mu$.
Another way I look at it is from a simulation point of view. Assuming I have complete access to the distribution of $X$ (simulating $Y$ is easy), I can simulate coupled $(X_1,Y_1)$ with $\mathbb P(X_1=1)\geq p$. Then simulate $X_2$ conditioned on the realized value of $X_1$ (which obviously has positive probability for nontrivial $p$), etc. At every step we must have the probability that $X_n=1$ is at least $p$ (conditioned on an event of positive probability). Of course, maybe the structure of $X$ could possibly prevent this type of simulation? Though as long as this is possible, then (\ref{eq1}), for all $n,m,\epsilon$, should be necessary and sufficient for $\pi_p\preceq \mu$.
Am I misunderstanding something?
Liggett & Steif (2006) explores the situation when $X$ has certain associations, and there are many other papers exploring special cases for certain types of processes as well.
Primary References:
Liggett et al (1997) https://www.jstor.org/stable/2959530
Liggett & Steif (2006) https://dx.doi.org/10.1016/j.anihpb.2005.04.002
(and follow any references from those and papers that cite them)
